(this post assumes you are in the UK, elsewhere YMMV)
This is usually caused by poorly terminated wires which leads to the terminals and wire being overheated.
The problem is overheating doesn't just tarnish the surface, it can also change the physical properties of the wire. Cleaning the tarnishing off is not enough the wire MUST be cut back. The terminations must be on fresh wire that has not been overheated and they must be tight. Failure to do this will almost certainly result in a repeat of the problem.
If there is not enough slack in the wiring then that needs to be dealt with by replacing wiring. Something like this 60A junction box with double screw terminations would be appropriate would be appropriate if you need to join cables but if the cable can be easilly replaced completely then that would be preferable. Remember junction boxes must remain accessible for inspection.
I would advise moving up to a 50A switch if possible to give a bit more headroom.
Personally I think the whole design of accessories for cooker and shower circuits in the UK is one of the less well designed parts of our wiring system. Fitting the accessories onto their backboxes requires moving the wiring which with fat inflexible cores often means some stress (especially if the wall is solid and the cables are direct burried so they can't move).
There is no law in the UK which would prohibit DIYing this work but if you are at all unsure of your abilities I would advise using the services of an electrician.
P.S. a note on power, voltage and current because a lot of people get this wrong.
Many people have heard the equation I=P/V and try to apply it in circumstances where it doesn't apply. Specifically the headline power figure for a resistive load (like a shower) will only be valid at one voltage. For UK electric showers the headline power is usually specified at 240V (despite the fact that our mains electricity is nominally 230V).
So to caclulate the current at a given voltage you would first calculate the resistance from the headline power and the voltage at which the headline power is specified (using R=V2/P). Then you can calculate the current from the resistance (which we can assume is constant over a reasonable range of voltages) and the voltage you want to know the corresponding current for.